A 4700 truck carrying a 900 kg crate is traveling at 25 m/s toright on straight

Question

A 4700 truck carrying a 900 kg crate is traveling at 25 m/s toright on straight level highway. Truck driver applies brakes,and as it slows down, the truck travels 55m in next 3.0 s. The crate doen't slide on back of truck a. Calculate manitude of acceleration of truck, assumin it isconstant. Note problem doesn't say truck comes to stop b. Calculate minimum coefficient of friction between crate andtruck that prevents crate from sliding c. Is thsi static or kinetic friction Assume bed of truck is frictionless but there is springconstant 9200 N/m attaching crate to truck. Truck is initallyat rest If truck and crate have same acceleration calculate extensionof spring as truck accelerates from rest to 25m/s in 10 s At some later time truck is moving at constatn speed of 25 m/sand crate is in equilibrium. Indicate whether extension ofspring is greater than, less than or same as part when truck wasaccelerationg A 4700 truck carrying a 900 kg crate is traveling at 25 m/s toright on straight level highway. Truck driver applies brakes,and as it slows down, the truck travels 55m in next 3.0 s. The crate doen't slide on back of truck a. Calculate manitude of acceleration of truck, assumin it isconstant. Note problem doesn't say truck comes to stop b. Calculate minimum coefficient of friction between crate andtruck that prevents crate from sliding c. Is thsi static or kinetic friction Assume bed of truck is frictionless but there is springconstant 9200 N/m attaching crate to truck. Truck is initallyat rest If truck and crate have same acceleration calculate extensionof spring as truck accelerates from rest to 25m/s in 10 s At some later time truck is moving at constatn speed of 25 m/sand crate is in equilibrium. Indicate whether extension ofspring is greater than, less than or same as part when truck wasaccelerationg

Explanation / Answer

(a) the average speed is    55/3 =    18.333 m/s . This is the average of initial and final speed,or     18.333  = (1/2) ( 25+ final speed) .    final speed =   11.6667 m/s . acc = change in speed / time = (25 - 11.6667 )/ 3   =   4.444m/s2 . (b) static friction will prevent the crate from sliding,and     max static friction force = u mg .       max acceleration =   max static friction force / m .      max acc =   ug .      u = acc / g = 4.444 / 9.8 =   0.454    is the minimum coeff of static frictionnecessary . (c) it is static friction. . (b) static friction will prevent the crate from sliding,and     max static friction force = u mg .       max acceleration =   max static friction force / m .      max acc =   ug .      u = acc / g = 4.444 / 9.8 =   0.454    is the minimum coeff of static frictionnecessary . (c) it is static friction.

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